This is more of a mathematical/conceptual question. I am reading through Chapter 12 of Shankar's "Principle's Of Quantum Mechanics" 2nd Edition, and in the very first section he describes a "consistency test" for the translation operator in that:

$$T(\mathbf{b})T(\mathbf{a})=T(\mathbf{a+b}) \tag{12.1.6}$$

$$ e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar }e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar }=e^{-i\mathbf{(a+b)}\cdot\mathbf{P}/\hbar } \tag{12.1.7}$$

Should be true and without proof, he says yes because the operators commute:

$$[P_x,P_y]=0 \tag{12.1.8}$$

This is what I don't understand, how would the exponentials combine if $[P_x,P_y]=0$ did not commute? It seems like they would combine in the same exact way. For example, lets suppose we have two translations in the $\vec{x}$ direction so that we have:

\begin{align} T(\mathbf{b\hat{x}})T(\mathbf{a\hat{x}})&=e^{-i\mathbf{b}\cdot\mathbf{P}/\hbar }e^{-i\mathbf{a}\cdot\mathbf{P}/\hbar }\\ &=e^{-ib{P_x}/\hbar }e^{-ia{P_x}/\hbar }\\ &=e^{-ib{P_x}/\hbar-ia{P_x}/\hbar}\\ &=e^{-\frac{i}{\hbar} (bP_x+aP_x) }\\ &=e^{-\frac{i}{\hbar} (b+a)P_x }\\ \end{align}

Which still satisfies the "law of combination" for the translation and does not even rely on commutativity. Can some one clear this up for me?


2 Answers 2


If $A$ and $B$ do not commute, then generally $$e^A e^B \neq e^{A+B}.$$ Then if $P_x$ and $P_y$ don't commute, we generally have $$T(a \hat{x}) T(b \hat{y}) \neq T(a \hat{x} + b \hat{y}).$$ You only considered the case of combining translations along the same direction; these will always compose nicely, basically by definition.

Clarification: let's expand both $e^A e^B$ and $e^{A+B}$ and compare the results. On the left-hand side, $$(1+A+A^2/2 + \ldots)(1+B+B^2/2 + \ldots) = 1 + (A + B) + \left(AB + \frac{A^2}{2} + \frac{B^2}{2}\right) + \ldots.$$ On the right-hand side, we have $$1 + (A+B) + \frac{(A+B)^2}{2} + \ldots.$$ The second order terms are not the same! The cross terms are $AB$ on the left-hand side, and $(AB+BA)/2$ on the right-hand side. So they're only generally equal if $A$ and $B$ commute.

  • $\begingroup$ I don't see how $$e^A e^B \neq e^{A+B}.$$ is true if A and B do not commute. Can you provide some explanation? $\endgroup$ Feb 15, 2017 at 21:23
  • $\begingroup$ @DonkeyKong I added a bit of explanation. $\endgroup$
    – knzhou
    Feb 15, 2017 at 21:30
  • $\begingroup$ See also physics.stackexchange.com/questions/132886/… $\endgroup$
    – gj255
    Feb 15, 2017 at 21:36

In general, for operators $X$ and $Y$ which do not necessarily commute, one has,

$$\ln(\exp X \exp Y) = X + Y + \frac12 [X,Y] + \frac{1}{12} \left([X,[X,Y]]+[Y,[Y,X]] \right) + \dots$$

which is known as the Baker-Campbell-Hausdorff formula. As you can see, if $P_x$ and $P_y$ commute, the higher order terms vanish and indeed the resultant expression is simply $P_x+ P_y$.

As a somewhat ad hoc example, one would have $e^a e^{a^\dagger} = e^{a + a^\dagger + \frac12}$ since $[a,a^\dagger]= 1$, and only the first commutator contributes, since higher terms are nested commutators and obviously constants commute with $a$ and $a^\dagger$.


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